The Ka value for ethanoic acid is 1.8x10^-5, what would be the pH of a 2.0x10^-2 M solution of ethanoic acid?
............CH3COOH ==> H^+ + CH3COO^-
initial.....0.02M........0.......0
change........-x.........+x......+x
final......0.02-x.........x.......x
Ka = (H^+)(CH3COO^-)/((CH3COOH)
Plug the ICE chart values into the Ka expression, solve for H, then convert to pH.